Let \(\displaystyle\theta={\frac{{{2}\pi}}{{{5}}}}\). Show \(\displaystyle{2}{\cos{{\left({2}\theta\right)}}}+{2}{\cos{{\left(\theta\right)}}}+{1}={0}\).

nomadzkia0re

nomadzkia0re

Answered question

2022-03-22

Let θ=2π5.
Show 2cos(2θ)+2cos(θ)+1=0.

Answer & Explanation

kattylouxlvc

kattylouxlvc

Beginner2022-03-23Added 11 answers

If 5θ=2π, 3θ=2π2θ
cos3θ=cos(2π2θ)=cos2θ
Using Triple and Double angle formulas,
4c33c=2c214c32x23c+1=0 (1)
where c=cosθ
Again, cos3θ=cos2θ
3θ=2nπ±2θ where n is any integer
Taking the '+' sign, 3θ=2nπ+2θθ=2nπ
cosθ=cos2nπ=1
Taking the '-' sign 3θ=2nπ2θθ=2nπ5 where n0,1,2,3,4
Again as cos(2πy)=cosy, cos2π5=cos(2π2π5)
=cos8π5
and cos4π5=cos6π5
Clearly, c1 is a factor of (1) and cos2π5, cos4π51
So, cos2π5=cos8π5 and cos4π5=cos6π5 are the roots of
4c32c23c+1c1=04c2+2c1=0
Again use Double formula cos2θ=2cos2θ1c2=cos2θ=1+cos2θ2
Kailee Castro

Kailee Castro

Beginner2022-03-24Added 8 answers

I think this is a simple way. By using De Moivre’s formula, we obtain the fifth time angle formula:
cos(5θ)=16cos5θ20cos3θ+5cosθ
Set θ=2π5, we have:
cos(2π)=16cos5θ20cos3θ+5cosθ
16cos5θ20cos3θ+5cosθ1=0
By using factorization, we have:
(cosθ1)(4cos2θ+2cosθ1)2=0
Obviously, cosθ10, then the equality have been prove.

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